Orthogonality of Bessel functions

Question

The orthogonality for Bessel functions is given by $\int_0 ^1 rJ_n(k_1r)J_n(k_2r) dr=0,\ (k_1 \neq k_2)\\ \neq 0, (k_1=k_2,\ J_n(k_1)=J_n(k_2)=0\ \mbox{or}\ J'_n(k_1)=J'_n(k_2)=0)$

This suggests a particular condition at the boundary $r=1$ for this orthogonality to hold. In case we have a different boundary condition for which $J_n(k_1,k_2) \neq 0\ \mbox{and}\ J_n'(k_1,k_2) \neq 0$, how do we establish an orthogonality condition for this case? How can we approach this problem?

This post imported from StackExchange Mathematics at 2014-06-02 20:22 (UCT), posted by SE-user vijay

Answer 1

$y=J_n(kr)$ is a solution of Bessel´s differential equation $r^2y'' +ry'+(r^2k^2-n^2)y=0$, which can be rewritten as $(ry')'+(rk^2-n^2/r)y = 0$.

If $u=J_n(ar)$ and $v=J_n(br)$, then they fulfill the equations

$$(ru')'+(ra^2-n^2/r)u = 0$$ $$(rv')'+(rb^2-n^2/r)v = 0$$

Multiply the first by $v$, the second by $u$ and substract them, and you get

$$(b^2-a^2)ruv =u(rv')'-v(ru')'=(vru'-urv')'$$

Integrating this, you get that

$$(b^2-a^2)\int_0^1ruvdr = \left.(vru'-urv')\right|_0^1=v(1)u'(1)-u(1)v'(1)$$

So if you want the left hand side to be $0$, then the right hand side must be $0$ as well, so you must have $aJ_n(b)J_n'(a)=bJ_n(a)J_n'(b)$.

This is fulfilled if $J_n(a)=J_n(b)=0$, or $J_n'(a)=J_n'(b)=0$, but also if $a J_n'(a)/J_n(a)=b J_n'(b)/J_n(b)$. So the boundary condition $y' = C y$ at $r=1$ will also work.

This post imported from StackExchange Mathematics at 2014-06-02 20:22 (UCT), posted by SE-user Jaime